Solution:

If the angle $\theta$ is with the vertical, then we need to adjust the force components accordingly:

1. Normal Force:

   $$N = mg - F \cos \theta$$

2. Friction Force:

   $$f_s = \mu N = \mu (mg - F \cos \theta)$$

3. Horizontal Component of Force:

   $$F_x = F \sin \theta$$

At the verge of motion:

$$F \sin \theta = \mu (mg - F \cos \theta)$$

Solving for $F$:

$$F = \frac{\mu mg}{\sin \theta + \mu \cos \theta}$$